A die is rolled, what is the probability of rolling a number that is greater than 6

Video transcript

Find the probability of rolling doubles on two six-sided dice numbered from 1 to 6. So when they're talking about rolling doubles, they're just saying, if I roll the two dice, I get the same number on the top of both. So, for example, a 1 and a 1, that's doubles. A 2 and a 2, that is doubles. A 3 and a 3, a 4 and a 4, a 5 and a 5, a 6 and a 6, all of those are instances of doubles. So the event in question is rolling doubles on two six-sided dice numbered from 1 to 6. So let's think about all of the possible outcomes. Or another way to think about it, let's think about the sample space here. So what can we roll on the first die. So let me write this as die number 1. What are the possible rolls? Well, they're numbered from 1 to 6. It's a six-sided die, so I can get a 1, a 2, a 3, a 4, a 5, or a 6. Now let's think about the second die, so die number 2. Well, exact same thing. I could get a 1, a 2, a 3, a 4, a 5, or a 6. Now, given these possible outcomes for each of the die, we can now think of the outcomes for both die. So, for example, in this-- let me draw a grid here just to make it a little bit neater. So let me draw a line there and then a line right over there. Let me draw actually several of these, just so that we could really do this a little bit clearer. So let me draw a full grid. All right. And then let me draw the vertical lines, only a few more left. There we go. Now, all of this top row, these are the outcomes where I roll a 1 on the first die. So I roll a 1 on the first die. These are all of those outcomes. And this would be I run a 1 on the second die, but I'll fill that in later. These are all of the outcomes where I roll a 2 on the first die. This is where I roll a 3 on the first die. 4-- I think you get the idea-- on the first die. And then a 5 on the first to die. And then finally, this last row is all the outcomes where I roll a 6 on the first die. Now, we can go through the columns, and this first column is where we roll a 1 on the second die. This is where we roll a 2 on the second die. So let's draw that out, write it out, and fill in the chart. Here's where we roll a 3 on the second die. This is a comma that I'm doing between the two numbers. Here is where we have a 4. And then here is where we roll a 5 on the second die, just filling this in. This last column is where we roll a 6 on the second die. Now, every one of these represents a possible outcome. This outcome is where we roll a 1 on the first die and a 1 on the second die. This outcome is where we roll a 3 on the first die, a 2 on the second die. This outcome is where we roll a 4 on the first die and a 5 on the second die. And you can see here, there are 36 possible outcomes, 6 times 6 possible outcomes. Now, with this out of the way, how many of these outcomes satisfy our criteria of rolling doubles on two six-sided dice? How many of these outcomes are essentially described by our event? Well, we see them right here. Doubles, well, that's rolling a 1 and 1, that's a 2 and a 2, a 3 and a 3, a 4 and a 4, a 5 and a 5, and a 6 and a 6. So we have 1, 2, 3, 4, 5, 6 events satisfy this event, or are the outcomes that are consistent with this event. Now given that, let's answer our question. What is the probability of rolling doubles on two six-sided die numbered from 1 to 6? Well, the probability is going to be equal to the number of outcomes that satisfy our criteria, or the number of outcomes for this event, which are 6-- we just figured that out-- over the total-- I want to do that pink color-- number of outcomes, over the size of our sample space. So this right over here, we have 36 total outcomes. So we have 36 outcomes, and if you simplify this, 6/36 is the same thing as 1/6. So the probability of rolling doubles on two six-sided dice numbered from 1 to 6 is 1/6.

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Sum

If one die is rolled, then find the probability of event that the number on the upper face is greater than 6?

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Solution

Sample space,

S = {1, 2, 3, 4, 5, 6}

∴ n(S) = 6

Let A be the event that the number on the upper face is greater than 6.

Here, the greatest number is 6.

∴ Event A is an impossible event.

∴ A = { }

∴ n(A) = 0

∴ P(A) = `("n"("A"))/("n"("S"))`

∴ P(A) = `0/6` = 0

Concept: Probability of an Event

  Is there an error in this question or solution?

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Chapter 5: Probability - Q.2 (B)

Q 3Q 2Q 4

APPEARS IN

SCERT Maharashtra Question Bank 10th Standard SSC Mathematics 1 Algebra Maharashtra State Board 2021

Chapter 5 Probability
Q.2 (B) | Q 3

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